= Solution
Here $t_0=0$ and $t_2=t_1^2$, so the constant and quadratic terms vanish, while the linear coefficient remains $t_1\ne0$. A further <derivative> gives
$$
f^{(3)}(x)=\frac{t^{(3)}(x)}{1+t(x)}-\frac{3t'(x)t''(x)}{(1+t(x))^2}+\frac{2t'(x)^3}{(1+t(x))^3}.
$$
Consequently $f^{(3)}(0)=t_3-3t_1t_2+2t_1^3=t_3-t_1^3$. Since $t_3$ and $t_1$ have opposite signs, $t_3$ and $t_1^3$ also have opposite signs, and their difference is nonzero. The first two nonzero terms are therefore
$$
\boxed{f(x)=t_1x+\frac{t_3-t_1^3}{6}x^3+O(x^4).}
$$
This is where the full sign condition $t_3t_1<0$ rules out a further cancellation in the <Taylor expansion>.
Back to article page
